$\overline{AB} = \sqrt{145}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $\sqrt{145}$ $?$ $ \sin( \angle BAC ) = \frac{9\sqrt{145} }{145}, \cos( \angle BAC ) = \frac{8\sqrt{145} }{145}, \tan( \angle BAC ) = \dfrac{9}{8}$
Solution: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{\sqrt{145}} $ $ \overline{AC}=\sqrt{145} \cdot \cos( \angle BAC ) = \sqrt{145} \cdot \frac{8\sqrt{145} }{145} = 8$